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Question: \((m + 2)\sin\theta + (2m - 1)\cos\theta = 2m + 1,\) i...

(m+2)sin⁡θ+(2m−1)cos⁡θ=2m+1,(m + 2)\sin\theta + (2m - 1)\cos\theta = 2m + 1, i

A

tan⁡θ=34\tan\theta = \frac{3}{4}

B

tan⁡θ=43\tan\theta = \frac{4}{3}

C

tan⁡θ=2mm2+1\tan\theta = \frac{2m}{m^{2} + 1}

D

None of these

Answer

tan⁡θ=43\tan\theta = \frac{4}{3}

Explanation

Solution

Squaring the given relation and putting tan⁡θ=t,\tan\theta = t,

(m+2)2t2+2(m+2)(2m−1)t+(2m−1)2=(2m+1)2(1+t2)(m + 2)^{2}t^{2} + 2(m + 2)(2m - 1)t + (2m - 1)^{2} = (2m + 1)^{2}(1 + t^{2})

⇒3(1−m2)t2+(4m2+6m−4)t−8m=0\Rightarrow 3(1 - m^{2})t^{2} + (4m^{2} + 6m - 4)t - 8m = 0

⇒(3t−4)[(1−m2)t+2m]=0\Rightarrow (3t - 4)\lbrack(1 - m^{2})t + 2m\rbrack = 0,

which is true if t=tan⁡θ=43t = \tan\theta = \frac{4}{3} or tan⁡θ=2mm2−1\tan\theta = \frac{2m}{m^{2} - 1}.